Difference between revisions of "2016 AMC 12A Problems/Problem 24"

Revision as of 20:31, 5 February 2016

Problem

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$ ?

$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution

The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value. Derive $f(x)$ so that the acceleration $f''(x)=0$ : $x^3-ax^2+bx-a\rightarrow 3x^2-2ax+b\rightarrow 6x-2a\rightarrow x=\frac{a}{3}$ for the inflection point/root. The function with the minimum $a$ :

$f(x)=\left(x-\frac{a}{3}\right)^3$ $x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}$ Since this is equal to the original equation $x^3-ax^2+bx-a$ ,

$\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}$ $b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}$

The actual function: $f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$

$f(x)=0\rightarrow x=\sqrt{3}$ triple root! "Complete the cube."

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 <cmath>\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}</cmath>
-<cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(C) }9}</cmath>
+<cmath>b=\frac{a^2}{3}=\frac{27}{3}=\boxed{\textbf{(B) }9}</cmath>
 The actual function:

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Difference between revisions of "2016 AMC 12A Problems/Problem 24"

Revision as of 20:31, 5 February 2016

Problem

Solution

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